Difference between revisions of "2020 AMC 8 Problems/Problem 24"
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+ | ==Problem 24== | ||
A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math> | A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math> | ||
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==Solution 1== | ==Solution 1== | ||
− | + | The area of the shaded region is <math>(24s)^2</math>. To find the area of the large square, we note that there is a <math>d</math>-inch border between each of the <math>23</math> pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of <math>23+2 = 25</math> times the length of the border, i.e. <math>25d</math>. Adding this to the total length of the consecutive squares, which is <math>24s</math>, the side length of the large square is <math>(24s+25d)</math>, yielding the equation <math>\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}</math>. Taking the square root of both sides (and using the fact that lengths are non-negative) gives <math>\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}</math>, and cross-multiplying now gives <math>120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}</math>. | |
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− | == | + | Note: Once we obtain <math>\tfrac{24s}{24s+25d} = \tfrac{4}{5},</math> to ease computation, we may take the reciprocal of both sides to yield <math>\tfrac{24s+25d}{24s} = 1 + \tfrac{25d}{24s} = \tfrac{5}{4},</math> so <math>\tfrac{25d}{24s} = \tfrac{1}{4}.</math> Multiplying both sides by <math>\tfrac{24}{25}</math> yields the same answer as before. ~peace09 |
− | + | ==Solution 2== | |
− | + | Without loss of generality, we may let <math>s=1</math> (since <math>d</math> will be determined by the scale of <math>s</math>, and we are only interested in the ratio <math>\frac{d}{s}</math>). Then, as the total area of the <math>576</math> gray tiles is simply <math>576</math>, the large square has area <math>\frac{576}{0.64} = 900</math>, making the side of the large square <math>\sqrt{900}=30</math>. As in Solution 1, the side length of the large square consists of the total length of the gray tiles and <math>25</math> lots of the border, so the length of the border is <math>d = \frac{30-24}{25} = \frac{6}{25}</math>. Since <math>\frac{d}{s}=d</math> if <math>s=1</math>, the answer is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>. | |
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− | ==Solution | + | ==Solution 3 (using answer choices)== |
− | + | As in Solution 2, we let <math>s = 1</math> without loss of generality. For sufficiently large <math>n</math>, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: | |
<asy> | <asy> | ||
draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); | ||
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} | } | ||
</asy> | </asy> | ||
− | Each red square has side length <math>1+d</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of area covered by the gray tiles | + | Each red square has side length <math>(1+d)</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of the total area covered by the gray tiles will be slightly less than <math>\frac{1}{(1+d)^2}</math>, which implies <math>\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}</math>. Hence <math>d</math> (and thus <math>\frac{d}{s}</math>, since we are assuming <math>s=1</math>) is less than <math>\frac{1}{4}</math>, and the only choice that satisfies this is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>. |
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+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/NHYB0VI3dcY | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solutions== | ||
+ | https://youtu.be/9tPTBxadV90 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=1515 | ||
+ | |||
+ | ~Interstigation | ||
==See also== | ==See also== |
Revision as of 22:27, 29 August 2021
Contents
Problem 24
A large square region is paved with gray square tiles, each measuring inches on a side. A border inches wide surrounds each tile. The figure below shows the case for . When , the gray tiles cover of the area of the large square region. What is the ratio for this larger value of
Solution 1
The area of the shaded region is . To find the area of the large square, we note that there is a -inch border between each of the pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of times the length of the border, i.e. . Adding this to the total length of the consecutive squares, which is , the side length of the large square is , yielding the equation . Taking the square root of both sides (and using the fact that lengths are non-negative) gives , and cross-multiplying now gives .
Note: Once we obtain to ease computation, we may take the reciprocal of both sides to yield so Multiplying both sides by yields the same answer as before. ~peace09
Solution 2
Without loss of generality, we may let (since will be determined by the scale of , and we are only interested in the ratio ). Then, as the total area of the gray tiles is simply , the large square has area , making the side of the large square . As in Solution 1, the side length of the large square consists of the total length of the gray tiles and lots of the border, so the length of the border is . Since if , the answer is .
Solution 3 (using answer choices)
As in Solution 2, we let without loss of generality. For sufficiently large , we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: Each red square has side length , so by solving , we obtain . The actual fraction of the total area covered by the gray tiles will be slightly less than , which implies . Hence (and thus , since we are assuming ) is less than , and the only choice that satisfies this is .
Video Solution by WhyMath
~savannahsolver
Video Solutions
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1515
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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